If the system of linear equations x + 2ay + az = 0, x + 3 by + bz = 0 , x + 4cy + cz = 0 has a non-zero solution, then a, b, c

Question

If the system of linear equations x + 2ay + az = 0, x + 3 by + bz = 0 , x + 4cy + cz = 0 has a non-zero solution, then a, b, c (A) are in A.P (B) are in G.P (C) are in H.P (D) satisfy a + 2b + 3c = 0

Tardigrade Admin · Accepted Answer

For non-zero solution |1&2a&a 1&3b&b 1&4c&c| = 0 ⇒ 3bc - 4bc - (2ac - 4ac) + 2ab - 3ab = 0 ⇒ - bc + 2 ac - ab = 0 ⇒ (2/b) = (1/a) + (1/c) ⇒ a, b, c are in H.P.

Q. If the system of linear equations $x + 2 a y + a z = 0, x + 3 b y + b z = 0, x + 4 cy + cz = 0$ has a non-zero solution, then a, b, c

are in A.P

are in G.P

are in H.P

satisfy $a + 2 b + 3 c = 0$

For non-zero solution $∣ ∣ 111 2 a 3 b 4 c a b c ∣ ∣ = 0$
$\Rightarrow 3 b c - 4 b c - (2 a c - 4 a c) + 2 ab - 3 ab$ = 0
$\Rightarrow - b c + 2 a c - ab$ = 0
$\Rightarrow \frac{2}{b} = \frac{1}{a} + \frac{1}{c} \Rightarrow a, b, c$ are in H.P.

Q. If the system of linear equations x+2ay+az=0,x+3by+bz=0,x+4cy+cz=0 has a non-zero solution, then a, b, c